Question

[Hoàng Thị Ngọc Mai]

Cho \(\left\{{}\begin{matrix}ab+bc+ca+abc=2\\a,b,c>0\end{matrix}\right.\)

Tìm Max :

\(P=\Sigma\dfrac{a+1}{a^2+2a+2}\)

@Akai Haruma

Answer 1

[Akai Haruma]

Lời giải:

Đặt \((a+1,b+1,c+1)=(x,y,z)\Rightarrow (a,b,c)=(x-1,y-1,z-1)\)

Khi đó:
\(ab+bc+ac+abc=2\)

\(\Leftrightarrow (x-1)(y-1)+(y-1)(z-1)+(z-1)(x-1)+(x-1)(y-1)(z-1)=2\)

\(\Leftrightarrow xyz-(x+y+z)+2=2\Leftrightarrow xyz=x+y+z\)

Vậy bài toán trở thành: Cho $x,y,z>0$ thỏa mãn \(x+y+z=xyz\)

Tìm max \(P=\sum \frac{x}{x^2+1}\)

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Ta có: \(x+y+z=xyz\Rightarrow x(x+y+z)=x^2yz\)

\(\Rightarrow x(x+y+z)+yz=yz(x^2+1)\)

\(\Leftrightarrow (x+y)(x+z)=yz(x^2+1)\Rightarrow x^2+1=\frac{(x+y)(x+z)}{yz}\)

Do đó: \(\frac{x}{x^2+1}=\frac{x}{\frac{(x+y)(x+z)}{yz}}=\frac{xyz}{(x+y)(x+z)}\)

\(\Rightarrow P=\sum \frac{x}{x^2+1}=\sum \frac{xyz}{(x+y)(x+z)}=\frac{2xyz(x+y+z)}{(x+y)(y+z)(x+z)}\)

Theo BĐT AM-GM:

\((x+y)(y+z)(x+z)=(x+y+z)(xy+yz+xz)-xyz\)

\(\geq (x+y+z).(xy+yz+xz)-\frac{(x+y+z)(xy+yz+xz)}{9}=\frac{8}{9}(x+y+z)(xy+yz+xz)\)

\(\Rightarrow P\leq \frac{2xyz(x+y+z)}{\frac{8}{9}(x+y+z)(xy+yz+xz)}=\frac{9}{4}.\frac{xyz}{xy+yz+xz}(*)\)

Mà: \((xy+yz+xz)^2\geq 3xyz(x+y+z)=3(xyz)^2\)

\(\Rightarrow xy+yz+xz\geq \sqrt{3}xyz(**)\)

Từ \((*);(**)\Rightarrow P\leq \frac{9}{4}.\frac{1}{\sqrt{3}}=\frac{3\sqrt{3}}{4}\). Vậy \(P_{\max}=\frac{3\sqrt{3}}{4}\)